Science:Math Exam Resources/Courses/MATH103/April 2013/Question 08 (c)

MATH103 April 2013

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Question 08 (c)
Fundamental Theorem of Calculus Find $\displaystyle \lim _{x\to 0}{\frac {1}{x^{3}}}\int _{0}^{x}{\frac {t^{2}}{1+t^{6}}}\,dt$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
If you plug in the endpoint of the limit (that is, $x=0$) into the numerator (the integral) and denominator, what value do you get? What technique can you use to solve this?

Hint 2
Try L'Hopital's rule.

Hint 3
(Alternate Solution 2) Try to integrate the integral.

Hint 4
Use a substitution.

Hint 5
Use the substitution $\displaystyle u=t^{3}$.

Hint 6
Then you can use L'Hopital's rule.

Hint 7
(Alternate Solution 3) Alternatively if you do not know L'Hopital's rule, you can use the Taylor series expansion for arctan.

Hint 8
(Alternate Solution 4) Try to figure out a Taylor series expansion for the integral and then use this to compute the limit.

Hint 9
Start with the expansion for $\displaystyle {\frac {1}{1-x}}$ and go from here.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Plugging in 0 into the integral yields $\int _{0}^{0}{\frac {t^{2}}{1+t^{6}}}\,dt=0$ and plugging in 0 into $\displaystyle x^{3}$ also yields 0 hence we may apply L'Hopital's rule to see that $\displaystyle \lim _{x\to 0}{\frac {1}{x^{3}}}\int _{0}^{x}{\frac {t^{2}}{1+t^{6}}}\,dt=\lim _{x\to 0}{\frac {\frac {x^{2}}{1+x^{6}}}{3x^{2}}}=\lim _{x\to 0}{\frac {1}{3(1+x^{6})}}={\frac {1}{3}}$ Notice that in the first equality, we also used the fundamental theorem of calculus.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let's try to evaluate the inside integral $\int _{0}^{x}{\frac {t^{2}}{1+t^{6}}}\,dt$. Let $\displaystyle u=t^{3}$ so that $\displaystyle du=3t^{2}dt$, $\displaystyle u(0)=(0)^{3}=0$ and $\displaystyle u(x)=x^{3}$. Plugging in yields $\int _{0}^{x}{\frac {t^{2}}{1+t^{6}}}\,dt=\int _{0}^{x^{3}}{\frac {{\tfrac {1}{3}}du}{1+u^{2}}}={\frac {1}{3}}\arctan(u){\big |}_{0}^{x^{3}}={\frac {\arctan(x^{3})}{3}}$. With the last equaltiy holding since plugging in 0 gives $\displaystyle {\frac {\arctan(0)}{3}}=0$. Now our goal is to evaluate the limit $\displaystyle \lim _{x\to 0}{\frac {\arctan(x^{3})}{3x^{3}}}$. Plugging in zero into the numerator and denominator gives an indeterminant form 0/0 and so we may apply L'Hopital's rule to see that $\displaystyle \lim _{x\to 0}{\frac {\arctan(x^{3})}{3x^{3}}}=\lim _{x\to 0}{\frac {\frac {3x^{2}}{1+x^{6}}}{3(3x^{2})}}=\lim _{x\to 0}{\frac {1}{3(1+x^{6})}}={\frac {1}{3}}$ completing the problem.

Solution 3

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Proceed as in solution 2 to see that the integral evaluates to $\displaystyle {\frac {\arctan(x^{3})}{3}}$. Now we seek to evaluate $\displaystyle \lim _{x\to 0}{\frac {\arctan(x^{3})}{3x^{3}}}$ Taking the Taylor expansion of the function yields ${\begin{aligned}{\frac {\arctan(x^{3})}{x^{3}}}&={\frac {1}{3x^{3}}}\sum _{n=0}^{\infty }(-1)^{n}{\frac {(x^{3})^{2n+1}}{2n+1}}\\&={\frac {1}{3x^{3}}}\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{6n+3}}{2n+1}}\\&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{6n}}{3(2n+1)}}\end{aligned}}$ As all terms bigger than when n equals 0 contain a positive power of x, taking the limit of the above as x tends to 0 leaves only the constant term which is $\displaystyle {\frac {(-1)^{0}x^{6(0)}}{3(2(0)+1)}}={\frac {1}{3}}$

Solution 4

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We start with the Taylor expansion for $\displaystyle {\frac {1}{1-x}}$ $\displaystyle {\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}$ Plugging in $\displaystyle x=-t^{6}$ yields $\displaystyle {\frac {1}{1+t^{6}}}=\sum _{n=0}^{\infty }(-t^{6})^{n}=\sum _{n=0}^{\infty }(-1)^{n}t^{6n}$ Multiplying by $\displaystyle t^{2}$ gives $\displaystyle {\frac {t^{2}}{1+t^{6}}}=\sum _{n=0}^{\infty }(-1)^{n}t^{6n+2}$ Now integrating from 0 to x gives $\displaystyle \int _{0}^{x}{\frac {t^{2}}{1+t^{6}}}=\sum _{n=0}^{\infty }(-1)^{n}\int _{0}^{x}t^{6n+2}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{6n+3}}{6n+3}}$ Dividing by $\displaystyle x^{3}$ gives $\displaystyle {\frac {1}{x^{3}}}\int _{0}^{x}{\frac {t^{2}}{1+t^{6}}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{6n}}{6n+3}}$ As all terms bigger than when n equals 0 contain a positive power of x, taking the limit of the above as x tends to 0 leaves only the constant term which is $\displaystyle {\frac {(-1)^{0}x^{6(0)}}{6(0)+3}}={\frac {1}{3}}$ which is the desired answer.

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